Question

[Kiều Vũ Minh Đức]

Chứng minh đẳng thức

\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-1}\) với x\(\ge\)2

Answer 1

[Trần Thanh Phương]

\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)

\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)

\(=\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1\)(*)

Vì \(x\ge2\Rightarrow x-1\ge1\Rightarrow\sqrt{x-1}\ge1\Rightarrow\sqrt{x-1}-1\ge0\)

Khi đó (*)\(=\sqrt{x-1}-1+\sqrt{x-1}+1=2\sqrt{x-1}\)(đpcm)